Chemistry 111 Sect. 2
Exam #2
April 11, 2002
KEY
Name: ________________________________________________
Student #: ______________________________________________
Please put your name and student number on all pages in case they become separated. The
last page of the exam has some useful information. The exam lasts 60 minutes. There are a
total of 100 points.
MULTIPLE CHOICE QUESTIONS
Place an X in the box corresponding to the correct answer.
1. (5 pts)
is a
†
X
†
†
†
†
The element X has an electron configuration of 1s2 2s2 2p6 3s2 3p4. This element
metal
non-metal
metalloid
transition metal
noble gas
There are 6 valence electrons above the Ne core, so this is a group
VI A element.
2. (5 pts) Which of the following uncharged atoms has the HIGHEST electronegativity χ?
†
†
†
†
X
†
Al
Cs
Na
P
S
Sulfur is farthest to the right and closest to the top of the periodic
table.
3. (5 pts) The bond order for the sulfur-oxygen bonds in the molecule SO2 is
†
X
†
†
†
†
1
1.5
2
2.5
3
..
:O
..
..
S
..
O:
The Lewis structure of sulfur dioxide has two resonance forms,
one of which is shown. There are two S-O links and three total
bonds (2 σ and 1 π), so the bond order is (1 + 2)/2 = 3/2 = 1.5.
Page 1
Chemistry 111 Sect. 2
Exam #2
April 11, 2002
KEY
Name: ________________________________________________
Student #: ______________________________________________
4. (5 pts) Estimate the enthalpy change ∆H for the following reaction. (Note: A table of bond
energies is given on the last page.)
C2H4 + 3 O2 → 2 CO2 + 2 H2O
X
†
†
†
†
†
-1032 kJ/mol
-848 kJ/mol
-561 kJ/mol
0 kJ/mol
+119 kJ/mol
There are 4 C-H and one C=C bond in C2H4, one O=O bond in O2, 2 C=O
bonds in CO2 and two H-O bonds in H2O. The sum of bond energies for the
reactants and products are
BEreact = 4×413 kJ/mol + 1×602 kJ/mol + 3×498 kJ/mol = 3748 kJ/mol
BEprod = 2×2×732 kJ/mol + 2×2×463 kJ/mol = 4780 kJ/mol
The enthalpy change is the sum of reactant bond energies minus the sum of
product bond energies:
∆H = 3748 kJ/mol – 4780 kJ/mol = -1032 kJ/mol. The reaction is exothermic.
5. (5 pts) Formaldehyde, H2CO, has the Lewis structure shown below. What is the
hybridization of the carbon atom?
†
X
†
†
†
†
sp
sp2
sp3
sp3d
sp3d2
There are 3 σ bonds on the carbon, so
three molecular hybrid orbitals are needed.
We use one s and two p atomic orbitals to
make the three sp2 hybrids. The π bond
H
comes from an unhybridized p orbital.
:O:
H
C
6. (5 pts) The partial Lewis structure of XeF3+ is shown below. The number of lone pair
electrons on the underlined Xe and F atoms in the complete Lewis structure is
(Xe, F)
†
†
†
†
X
†
0,6
0,7
1,6
2,0
2,6
The total number of valence electrons
equals 8 + 3×7 – 1 = 28. Six LP electrons
go on each F, leaving 4 electrons that make
two LP on the Xe.
The answers to the left were for # of LP on
the Xe and # of LP electrons on the F. It
was too complicated to explain this during
the exam, so full credit was given for any
answer.
Page 2
+
F
F
Xe
F
Chemistry 111 Sect. 2
Exam #2
April 11, 2002
KEY
Name: ________________________________________________
Student #: ______________________________________________
7. (5 pts) The carbonate ion, CO32-, has how many Lewis resonance structures?
†
†
†
X
†
†
0
One Lewis structure is shown at the right. The double
bond could also be formed from the lower right or lower
1
left O, so there are two other possible Lewis structures,
2
for a total of 3 resonance forms.
3
can’t tell from the information given
2-
:O:
.. C
:O
..
..
O:
..
8. (5 pts) Which of the following pairs of bonds is NOT shown as longer first, shorter
second?
(longer, shorter)
†
†
X
†
†
†
Ga—Cl, B—Cl
P—S, P—O
P=S, P—S
N—N, N≡N
Se—O, O—O
For the first two and last pair, one of the atoms in the left
pair is larger than the corresponding one on the second
pair, so the left bond is longer due to atom size. Single
bonds are longer than double are longer than triple if the
bonded pair of atoms are the same.
MULTIPLE MULTIPLE CHOICE QUESTIONS
Place an X in the boxes corresponding to the correct answers. Any number of answers may
be correct, including none of them.
9. (10 pts) Which of the following molecules is POLAR?
†
X
†
X
†
X
†
†
CH4
CH3Cl
CH2Cl2
CHCl3
CCl4
All of these molecules have tetrahedral electronic and
molecular geometries. The first and last are symmetrical,
so they won’t be polar. The middle three have polar
bonds and are not symmetrical, so they will be polar.
Page 3
Chemistry 111 Sect. 2
Exam #2
April 11, 2002
KEY
Name: ________________________________________________
Student #: ______________________________________________
10. (10 pts) Which of the following molecular geometries can arise when the central atom
has sp3d hybridization?
X
†
†
†
X
†
†
linear
trigonal planar
tetrahedral
see-saw
square planar
The electronic geometry for five sigma plus LP electron
pairs is trigonal bipyramidal. Having one LP and 4
terminal atoms makes a see-saw. Having 3 LPs and 2
terminal atoms (the LPs always add in the three planar
orbitals first) makes a linear molecule. The trigonal
planar arrangement won’t work because LPs go in the
three planar orbitals first.
11. (10 pts) Which of the following are TRUE?
X
†
X
†
X
†
X
†
†
A fluoride ion has a larger radius than a fluorine atom. The extra electron makes F- larger.
H2O is polar. Bent molecular geometry, polar bonds.
There are 36 core electrons for Sr. It has a Kr core and two electrons in 5s.
Cl has a greater electron affinity than I. Cl is above I in the periodic table.
The 5s orbital is lower energy than the 3p orbital.
SHORT ANSWER QUESTIONS
Answer in the spaces indicated.
b
12. (8 pts) The Lewis structure of the amino acid glycine is shown below.
H H :O:
sp3 because four orbitals are needed
a) The hybridization of N is _______________________________
:N C
a
C
..
O
.. H
H H
c
2
because three orbitals are needed for LP + σ
b) The hybridization of the double-bonded O is sp
_______________
109.5º because the N has tetrahedral electronic geometry
c) The C-N-H bond angle is _______________________________
109.5º because the O has tetrahedral electronic geometry
d) The C-O-H bond angle is _______________________________
Page 4
d
Chemistry 111 Sect. 2
Exam #2
April 11, 2002
KEY
Name: ________________________________________________
Student #: ______________________________________________
13. (8 pts) Draw the Lewis structure for PO43-.
P is the central atom. The total # of valence electrons is 5 + 4×6
+ 3 = 32. Make 4 σ bonds and put the remaining 24 electrons as
12 LP on the oxygens. All atoms have an octet, so that’s all that
needs to be done. This is the standard Lewis structure. An
alternate has one oxygen LP moved to a double bond, which
would be one of four possible resonance structure. This places
zero formal charge on the P and the double-bonded O. Full credit
for this structure was given. This is not the standard Lewis
structure since it leaves P with 10 electrons surrounding it.
..
:O
..
..
:O:
P
3-
..
O:
..
:O:
..
14. (8 pts) The Lewis structure for SF4 is given below. What is the molecular geometry of
SF4?
..
:F:
.. ..
..
:F
F:
S
..
..
:F:
..
The sulfur has five LP + σ bonds, so the electronic
geometry is trigonal bipyramidal. The one LP goes into
one of the three planar molecular orbitals that are 120º
apart, leaving four S-F bonds. Two are perpendicular to
the planar orbitals and two are in the planar orbitals.
This is a see-saw molecular structure.
14. (6 pts) What are the formal charges on each of the three atoms in NO2- in the Lewis
structure shown below?
-
Left O: fc = 6 – 1 – 6 = -1
..
:O
..
N : fc = 5 – 3 – 2 = 0
Right O: fc = 6 – 2 – 4 = 0
The negative charge on the ion is dominantly on the
leftmost oxygen for this structure. It is one of two
resonance structures. The negative formal charge is
always on one of the oxygens (the one that is singlebonded).
Page 5
..
N
..
O:
KEY
Name: ________________________________________________
Student #: ______________________________________________
Chemistry 111 Sect. 2
Exam #2
April 11, 2002
Useful information
NA = 6.02 × 1023 molecules/mol
d = m/V
c = 3.0 × 108 m/sec
E = -Rhc/n2
h = 6.6 × 10-34 J sec
Rhc = 2.2 × 10-18 J
me = 9.1 × 10-28 g
a0 = 0.0529 nm
2-linear, 3-trigonal planar, 4-tetrahedral
5-trigonal bipyramidal, 6-octahedral
E = hν
c = λν
λ = h/(m v)
Z* = Z - ninner electons
∆Hrxn = Σ(bond ener)react - Σ(bond ener)prod
Bond
H-C
H-H
H-N
H-O
C-C
Some useful data
Bond Bond
Bond
energy
(kJ/mol)
C-N
305
C=C
C-O
358
C≡C
N-N
163
C=N
N-O
201
C≡N
O-O
146
C=O
Bond
energy
(kJ/mol)
413
436
391
463
346
Bond
energy
(kJ/mol)
602
835
615
887
732
Bond
Bond
energy
(kJ/mol)
1072
418
945
607
498
C≡O
N=N
N≡N
N=O
O=O
PERIODIC TABLE OF THE ELEMENTS
IA
IIA
IIIB
IVB
VB
VIB
VIIB
VIIIB
IB
IIB
IIIA
IVA
VA
VIA
VIIA VIIIA
1
2
H
He
1.008
4.003
3
4
5
6
7
8
9
Li
Be
B
C
N
O
F
10
Ne
6.939
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.31
19
20
21
22
K
Ca
Sc
Ti
39.10
40.08
44.96
37
38
39
Rb
Sr
85.47
55
26.98
28.09
30.97
32.07
35.45
39.95
24
25
26
27
28
29
30
31
32
33
34
35
36
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
47.90
50.94
52.00
54.94
55.85
58.93
58.71
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
87.62
88.91
91.22
92.91
95.94
(99)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9
137.3
138.9
178.5
181.0
183.8
186.2
190.2
192.2
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
87
88
89
Fr
Ra
Ac
(223)
226.0
227.0
23
Page 6
I
Xe
Chemistry 111 Sect. 2
Exam #2
April 11, 2002
KEY
Name: ________________________________________________
Student #: ______________________________________________
Page 7